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Computers in Civil Engineering 53:081 Spring 2003. Lecture #6. Roots of Equations (Chapter 5 of 3rd Edition of C 2 ). Solve for x please …. 4x=8 (x-1)(x+1)=0 e -x =x cos(x)cosh(x)+1=0. Problem Formulation. Find roots of equations in form: f(x)=0.

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computers in civil engineering 53 081 spring 2003

Computers in Civil Engineering53:081 Spring 2003

Lecture #6

Roots of Equations(Chapter 5 of 3rd Edition of C2)

solve for x please
Solve for x please …
  • 4x=8
  • (x-1)(x+1)=0
  • e-x=x
  • cos(x)cosh(x)+1=0
problem formulation
Problem Formulation

Find roots of equations in form:f(x)=0

(We can put any of our one-equation engineering problems into this form)Question: What do we mean by roots?

Question: Why do we need to find the roots of such equations?

Answer: Example of uniform flow in open channel

slide4

Uniform Flow

Colebrook-White Formula

T(y)=surface width

y

A(y)=area

P(y) = wetted perimeter

continue
Continue...

In many instances Q, So, etc. are known, and one wants to solve for y. In other words: given Q and channel geometry, solve:

f(y) = 0

Problem: Explicit solution does not exist! (especially since A(y) and P(y) are also functions)

Solution:Numerically find root of equation, e.g. find value of depth y such that the equation is satisfied.

(Note that you can move Q to the right side and say 0=f(y). What value of y causes f(y) to be zero?)

outline
Outline

By definition f(xr) = 0

  • Single roots
  • Multiple roots
  • Systems of equations

fi(x1,x2,...,xn)=0 for i = 1,...,n

graphical method
Graphical Method
  • Plot function and observe zero crossing- ALWAYS do this first!
  • Useful for finding initial conditions of systematic methods

f(x)

x

example
Example

Consider the function:

i.e.

or

the graphical approach
The Graphical Approach

1.0

0.5

f(x)

0.0

0.2

0.4

0.6

0.8

-0.5

-1.0

x

general discussion
General Discussion
  • Bracketing Methods
    • Require two initial guesses that bracket the true root
  • Open Methods
    • Require only one initial guess
  • Special case: multiple roots. For example:

Two roots at 1

bracketing methods
Bracketing Methods
  • Bisection Method
  • False-Position Method (Regula-Falsi)
possible situations
Possible Situations

f(x)

f(x)

f(xu)

f(xl)

x

x

f(xl)f(xu) < 0 Odd number of roots in bracket

f(xl)f(xu) > 0 No root in bracket

f(x)

f(x)

x

x

f(xl)f(xu) > 0 Even number of roots in bracket

f(xl)f(xu) < 0 Odd number of roots in bracket

difficult situations
Difficult Situations

(f(xl)f(xu)<0 but even number of roots

f(x)

f(x)

f(xlu)

x

x

f(xl)

the bisection method
The Bisection Method

1. Choose lower xl and upper xu bounds.

2. Check if f(xl) f(xu) < 0. If not, go to 1.

3. Estimate root by xr = 0.5(xl+xu)

4. Check which interval contains the root.

If f(xl) f(xr) < 0 set xu = xr and go to 5.

If f(xl) f(xr) > 0 set xl = xr and go to 5.

If f(xl) f(xr) = 0 stop. (is this possible?)

5. Is the required accuracy met?

If yes: Stop.

If not: Go to 3.

termination criteria
Termination Criteria

Iteratively estimate error:

In the present context:

Where is the prescribed stopping criterion.

example f x e x x
Example: f(x) = e-x -x

1.0

By definition f(xr) = 0

0.5

f(x)

0.0

0.2

0.4

0.6

0.8

-0.5

-1.0

xu

xl

x

termination criteria1
Termination Criteria

Iteration xr|t|% |a|%

1 0.5 11.8

2 0.75 32.2 33.3

3 0.625 10.2 20.0

4 0.5625 0.819 11.1

5 0.59375 4.69 5.3

(The root is approximately at 0.567)

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